It occurs to me that at some point John might want to reconsider the initial principle about Mary’s inverse intelligence and start wearing that one down by degrees in the same way. But that possibility isn’t allowed for in the original argument so I don’t see why it should be here.

Or, what if the point at issue is not moonwalking but instead whether Mary is inversely intelligent. How can John decide the principle without using it to make his decision? I don’t know that he can and since the answer to the question about Mary’s intelligence is an assumption upon which the argument relies it really has to be the first one answered before we can figure out moonwalking.

Imagine that you are one of these people, and are aware of only 3 other blue eyed people. This means you might be in a population with only 3 blue eyed people out of 100, or you might be in a population with 4 blue eyed people, one of whom is you. How can you tell the difference between these two possibilities?

For silence to convey that information, you have to assume that people learn in discrete time quanta that are *synchronous* with each other, and furthermore, that they all *know* that their time quanta of fact-learning are synchronous. This is what would allow you to figure out, after three discrete “learning” steps, that the fact that each of the 3 blue eyed people you observe hasn’t acted yet means that they each also observe 3 blue eyed people, because if they each observed only two, they would all have acted one time unit earlier.

But if you don’t assume synchrony and discrete-ness, you have know way of knowing how quickly each person observes other people’s observations relative to how quickly you observe them, or relative to how quickly anyone else does, so you have no basis for deciding when the end condition of this mutual learning has been reached. Furthermore, you know that nobody else has any basis for knowing when this end condition should be reached, either. What if those other 3 blue eyed people haven’t killed themselves yet simply because they don’t yet know that each of the other 3 hasn’t yet decided that each of the other 3 hasn’t yet decide… and you might be brown-eyed after all. And merely having that doubt means that you know that all 3 of them should have the same doubt, so that even if you are blue-eyed it wouldn’t give them reliable knowledge.

Otherwise, what’s to stop every single brown-eyed person from concluding that they’re blue-eyed?

Remember, we don’t even need to assert that people *don’t* learn discrete facts one by one on a synchronized universal clock. All we need to assert is that, in the (unlikely) case that they do, they don’t *all* *know* that everyone learns the same kinds of discrete facts one by one on a synchronized universal clock.

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To say, “If Moe knows (1), then Moe knows (2),” is not the same as saying “Moe knows (1) is the same as Moe knowing (2)”. Thus, when I correct your

(1) Moe knows “Larry knows there are at least two blue eyed people on the island (namely Shemp and Curly)”

to

(1′) Moe knows “Either Larry knows
(a) there are at least two blue-eyes (Shemp and Curly),
or Larry knows
(b) there are at least three blue-eyes (Shemp, Curly, and me)”,

I am actually making a substantive change to the content of Moe’s knowledge: He knows more than just that when Larry writes down his lower bound for blues, Larry writes down an N > 1; Moe knows that if Moe’s eyes are brown, Larry writes down N = 2, while if Moe’s eyes are blue, Larry writes down N = 3. Yes, this implies Larry’s N is at least 2; but Moe knows more than just that, he knows the circumstances under which Larry’s N is 2 and the circumstances under which Larry’s N is 3.

However, I don’t find that it makes any substantive difference to the conclusion. As I said, I still am unsettled in this matter. The stranger doesn’t say anything other than what anyone else might have said, had the taboo on discussion not been in place, so how is this additional information? Maybe I need to digest Kareem’s points with only 3 stooges.

Steve H.

There are always at least 0 blue-eyed people. Did you perhaps overlook the ‘at least’ part of the statement?

There are actually two statements, of particular relevance to my argument, that will be present in the population.

Let me invent some notation here. Let

“(Everybody knows that)^2 there are at least 0 blue-eyed people.”

mean:

“Everybody knows that Everybody knows that there are at least 0 blue-eyed people.”

Then, if I can see N blues, I can eventually conclude that:

(Everybody knows that)^N there are at least 0 blue-eyed people.

If there are N blue-eyed people in the population then browns see N blue-eyed people and blues see N-1 blue-eyed people. Thus, browns believe that:

(Everybody knows that)^N there are at least 0 blue-eyed people.

While, all blues (who can only see N-1 blue-eyed people) believe that:

(Everybody knows that)^(N-1) there are at least 0 blue-eyed people.

The stranger changes the situation so that:

(Everybody knows that)^N there is at least 1 blue-eyed person.

(Everybody knows that)^(N-1) there is at least 1 blue-eyed person.

So he gives both groups new information.

This then gives them a mechanism for counting the number of people by counting the number of days without suicide.

Once that information has been received, each person would start on the assumption that they are brown and pick any blue eyed person to observe their behavior and determine if they are blue eyed themselves. It doesn’t matter which they choose.